Digitale Zaubertinte: Steganography workshop

Author: Martin Beneš, Verena Lachner

github.com/uibk-uncover/mip-stego-demo

Let's play a coin game!

Source: https://giphy.com/

$1000$ people toss a coin $100$ times, and count heads.

import numpy as np
toss = np.random.choice([0, 1], size=[1000, 100])  # toss
toss_sums = np.sum(toss, axis=1)  # count heads

import seaborn as sns
sns.histplot(toss_sums, discrete=True);  # plot histogram

Histograms allows us to analyze data distribution graphically.

Playing with unfair coin

Source: https://giphy.com/

Our unfair coin yields heads with a probability $52\%$.

toss2 = np.random.choice([0, 1], size=[1000, 100], p=[.48, .52])
toss2_sums = np.sum(toss2, axis=1)

sns.histplot(toss_sums, discrete=True, label='0.5');  # fair
sns.histplot(toss2_sums, discrete=True, label='0.52');  # unfair
plt.legend();

Is the orange histogram same as the blue one?

We can test it using statistics.

$\chi^2$ test compares the shapes of two histograms, $h$ and $g$.

$$X^2=\sum_{i=1}^{N}\frac{(h_i - g_i)^2}{h_i}$$

# histograms
h = np.histogram(toss_sums, bins=range(100+1))[0]
g = np.histogram(toss2_sums, bins=range(100+1))[0]

# test statistics
test_stat = 0
for i in range(100):
    if h[i] > 0:
        test_stat += (h[i] - g[i])**2 / h[i]

test_stat  # test statistic

197.7241088701197

We convert the test statistic to p-value (using $\chi^2$ distribution).

scipy.stats.chi2.sf(test_stat, 100-1)  # significant if <0.05

1.4826819136900469e-08

The histograms are different. Something's wrong with the coin!

Enough of coins, back to images!

Source: https://giphy.com/

# load image
x = np.array(Image.open('nockspitze.png').convert('L'))
plt.imshow(x,cmap = "gray");

sns.histplot(x.flatten(), discrete=True);  # cover histogram

sns.histplot(lsbr(x, 1.).flatten(), discrete=True);  # stego histogram

Let's take a closer look. What is going on with the histogram?

fig, ax = plt.subplots(1, 3, sharey=True)
for i, alpha in enumerate([.0, .5, 1.]):
    y = lsbr(x, alpha).flatten()
    sns.histplot(y, binrange=(125, 175), discrete=True, ax=ax[i]);
    ax[i].set_title(f'{alpha:.1f}');

LSBr averages the neighbor pairs (even and odd neighbor).

$$\bar{h}_i=\frac{h_{i}+h_{i+1}}{2}$$

# histogram
h, edges = np.histogram(x.flatten(), bins=range(256+1))
# average even-odd pairs
hbar = np.repeat((h[:-1:2] + h[1::2]) / 2, 2)

fig, ax = plt.subplots(1, 2, sharey=True)
ax[0].bar(range(256), h);
ax[1].bar(range(256), hbar);

Can we use $\chi^2$ test to detect steganography?

If the histogram is similar to the pair-averaged histogram, steganography is present.

$$S=\sum_{i=0}^{255}\frac{(h_i-\bar{h}_i)^2}{\bar{h}_i}$$

# Avoid division by zero
h = h[hbar > 0]
hbar = hbar[hbar > 0]
# Chi2 test
test_stat = np.sum((h - hbar)**2 / hbar)
pvalue = scipy.stats.chi2.sf(test_stat, 2**8-1)

pvalue  # stego if >0.05

0.0

P-value is less than $5\%$, histograms are different. The image is cover.

We run the $\chi^2$-test for stego.

def chi2_attack(x):
    # histograms
    h = np.histogram(x.flatten(), bins=range(256+1))[0]
    hbar = np.repeat((h[:-1:2] + h[1::2])/2, 2)
    h, hbar = h[hbar > 0], hbar[hbar > 0]
    # chi2 test
    S = np.sum((h[:-1:2] - hbar[::2])**2 / hbar[::2])
    return scipy.stats.chi2.sf(S, h.size-1)

y = lsbr(x, 1.)  # create stego
pvalue = chi2_attack(y)  # chi2 attack

pvalue  # stego if >0.05

1.0

P-value is greater than $5\%$, histograms are the same. The image is stego.

Take-away messages

• Steganography distorts image statistics.

• $\chi^2$ test can detect the presence of LSB replacement steganography.

Hands-on: LSBr

• Find the image(s) with steganography among the suspicious images
• Try to extract the messages
  • The secret key is the answer to life, the universe, and everything.

# load suspicious images
t21a = np.array(Image.open(f't21a.png'))
martinswand = np.array(Image.open(f'martinswand.png'))
statue = np.array(Image.open(f'statue.png'))

Run $\chi^2$ test on each image.

chi2_attack(t21a)

0.0

chi2_attack(statue)

0.0

chi2_attack(martinswand)

1.0

key = 42  # answer to life
message = extract_lsbr(martinswand, key=key)

print(message[:300], '...')

1609

THE SONNETS

by William Shakespeare



                     1
  From fairest creatures we desire increase,
  That thereby beauty's rose might never die,
  But as the riper should by time decease,
  His tender heir might bear his memory:
  But thou contracted to thine own bright eyes,
  Feed's ...

print(extract_lsbr(t21a))  # hidden Easter egg

Hi Verena!
Sorry again that you lost the key from your bike.
Let's meet today by the bike stand to cut the lock.

Martin.